$ABCD$ is a rectangle, $P$ is the midpoint of $AB$, and $Q$ is the point on $PD$ such that $CQ$ is perpendicular to $PD$. Prove that the triangle $BQC$ is isosceles.

Clearly, we need to prove that $BC=BQ$ or $\measuredangle{BCQ}=\measuredangle{CQB}$.

After a bit of angle chasing, a bit of Pythagoras, a bit of trig, I'm still not very close to proving either. I have an expression for $BQ$ using the cosine rule, but it's pretty messy and I'm unsure I can unscramble the mess and make it equal $BC$.

If anyone could come up with a proof, or at least a hint, that would be much appreciated.

  • $\begingroup$ There's a nice proof of this using vectors. Have you tried using vectors? $\endgroup$ – David Quinn Jul 20 '15 at 17:10
  • $\begingroup$ Hmmm, I'm not so comfortable with vectors, I can't see anything in particular, I was wondering if I may be able to exploit the fact that $CQ$ is perpendicular to $PD$, but I'm not sure. $\endgroup$ – MadChickenMan Jul 20 '15 at 17:25

Hint: draw a line from $B$ parallel to $DP$. This will intersect $CD$ at its midpoint and then...

  • $\begingroup$ Aha, I see now, I guess I was going after proving that $BC$ was equal to $BQ$ directly, too strongly, I didn't even consider trying to prove it like that. Lesson learnt. Thank you a lot. $\endgroup$ – MadChickenMan Jul 20 '15 at 18:07

This is an outline of the vector method:

Let $\overrightarrow{AB}=2\underline{a}$ and $\overrightarrow{BC}=\underline{b}$

Then $\overrightarrow{PQ}=\lambda(\underline{b}-\underline{a})$ and $\overrightarrow{CQ}=-\underline{b}-\underline{a}+\lambda(\underline{b}-\underline{a})$

Now using scalar product, we have $\overrightarrow{CQ}.\overrightarrow{PD}=0$

Recalling that $\underline{a}.\underline{b}=0$, you should get $$\lambda=\frac{b^2-a^2}{a^2+b^2}$$

Now work out $\overrightarrow{BQ}$ and then work out $\overrightarrow{BQ}.\overrightarrow{BQ}$ and you get $b^2$

And that's it.

  • $\begingroup$ Thanks for your solution. As I said, I have only done a bit of work on vectors, I couldn't understand why $\overrightarrow{PQ}=\lambda(b-a)$. Also, what calculations did you perform when making use of the scalar product, I only really know what to do when we're working in components. $\endgroup$ – MadChickenMan Jul 20 '15 at 18:39
  • $\begingroup$ PQ is parallel to PD and hence is a multiple of the vector PD. The dot product can be evaluated without using components : here all you need is $a.a=a^2$, $a.b=0$ because they're perpendicular, and you multipy out brackets in the usual way. $\endgroup$ – David Quinn Jul 20 '15 at 18:45
  • $\begingroup$ I get you now. Just one last thing though, so $\overrightarrow{BQ}=\overrightarrow{PQ}-a$? Doing the algebra, I think I get $\overrightarrow{BQ}=\frac{b^3-2ab^2-a^2b}{a^2+b^2}$, not the $b$ I was hoping for, have I gone wrong somewhere? $\endgroup$ – MadChickenMan Jul 20 '15 at 18:59
  • $\begingroup$ Sorry I'm being lazy with vector notation. The correct expression for $\overrightarrow{BQ}$ is $\frac{\underline{b}b^2-2b^2\underline{a}-a^2\underline{b}}{a^2+b^2}$ $\endgroup$ – David Quinn Jul 20 '15 at 19:07
  • $\begingroup$ Am I right in thinking the $ab$ terms cancel and we can say $b$ times by the denominator equals the numerator. Hence the required result? $\endgroup$ – MadChickenMan Jul 20 '15 at 19:38

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